This preview shows pages 1–3. Sign up to view the full content.
Lesson 5.5 Homework Solutions
1.
The resultant amplitude is given by:
2 cos
2 0.02 cos
2
12
0.039 mm
s
A
δ
π
=
= ×
×
=
2.
(i)
Resultant amplitude =
2 cos
2 0.05 cos
2
4
0.07 mm
A
= ×
×
=
(ii)
Resultant amplitude =
2 cos
2 0.05
0
cos
2
2
A
= ×
×
=
3.
(i)
1
340 ms
68
z
.
H
5 0
v
f
m
λ
=
=
=
(ii)
Path difference
∆
x = 11.2 – 10 = 1.2 m
1.2 m
5 m
0.24
x
∆
=
=
(iii)
The phase difference that corresponds to a path difference of
λ
= 2
π
Phase difference for 0.24
λ
=
0.48
π
(iv)
The resultant amplitude =
2 cos
2 0.01 cos0.2
0.015
4
2
m
A
m
= ×
×
=
4.
40.07
40
2.4
Path difference 40.07 – 40
= 0.07 m
(i)
The condition for minimum sound is:
(2
1)
0.07
2
n
+
=
For this condition, the frequency will be minimum when n = 0
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThen,
λ
/2 = 0.07,
or
λ
= 0.14 m
340
2429
0.14
v
f
Hz
λ
=
=
=
The condition for maximum sound is n
λ
= 0.07
The minimum frequency for this condition occurs when n = 1
Then
λ
= 0.07
340
0.07
4857
v
Hz
f
=
=
=
5.
442 Hz or 438 Hz
This is the end of the preview. Sign up
to
access the rest of the document.
 Fall '10
 George
 Physics, Work

Click to edit the document details